Đặt M = \(1+9+9^2+......+9^{2010}\)

\(9M=9+9^2+9^3+......+9^{2011}\)

\(9M-M=8M=9^{2011}-1\)

Đặt K = \(1+9+9^2+......+9^{2009}\)

\(9K=9+9^2+9^3+.....+9^{2010}\)

\(9K-K=8K=9^{2010}-1\)

\(\Rightarrow A=\frac{9^{2011}-1}{9^{2010}-1}\)

Đặt H=\(1+5+5^2+....+5^{2010}\)

\(5H=5+5^2+......+5^{2011}\)

\(5H-H=4H=5^{2011}-1\)

ĐẶT G = \(1+5+5^2+.......+5^{2009}\)

\(5G-G=4G=5^{2010}-1\)

\(\Rightarrow B=\frac{5^{2011}-1}{5^{2010}-1}\)

Rồi bạn so sánh sẽ ra ngay

A = \(1+\frac{9^{2010}}{1+9+9^2+....+9^{2009}}\)= \(1+1:\frac{1+9+9^2+....+9^{2009}}{9^{2010}}\)= \(1+1:\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+\frac{1}{9^{2008}}+...+\frac{1}{9}\right)\)

B = \(1+\frac{5^{2010}}{1+5+5^2+....+5^{2009}}\)= \(1+1:\frac{1+5+5^2+...+5^{2009}}{5^{2010}}\)= \(1+1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)

Do \(\frac{1}{9^{2010}}

có đúng đề không vậy 

Ta có :

+) \(A=\dfrac{1+9+9^2+...+9^{2009}}{1+9+9^2+...+9^{2009}}+\dfrac{9^{2010}}{1+9+9^2+...+9^{2009}}\)

\(A=1+1:\dfrac{1+9+9^2+...+9^{2009}}{9^{2010}}\)

\(A=1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)\)

+) \(B=\dfrac{1+5+5^2+...+5^{2009}}{1+5+5^2+...+5^{2009}}+\dfrac{5^{2010}}{1+5+5^2+...+5^{2009}}\)

\(B=1+1:\dfrac{1+5+5^2+...+5^{2009}}{5^{2010}}\)

\(B=1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)

Vì \(\dfrac{1}{9^{2010}}< \dfrac{1}{5^{2010}}\)

\(\dfrac{1}{9^{2009}}< \dfrac{1}{5^{2009}}\) (ngoặc cả mấy cài so sánh này vào rôi mời suy ra nhé)

.............................

\(\dfrac{1}{9}< \dfrac{1}{5}\)

\(\)=> \(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}< \dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\)

=> \(1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)

=> \(1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)

Hay A > B

ai làm được cho 10 tick

a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)

                 \(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B                                                                                        Vay A<B    

b,lam tuong tu nhu y a

Số số hạng của dãy:(2010-7):1+1=2004(số)

Vậy có tất cả:2004:2=1002(cặp)

A=7-8+9-10+11-12+...+2009-2010

A=(7-8)+(9-10)+(11-12)+...+(2009-2010)

A=-1+(-1)+(-1)+...+(-1)

Vậy A=(-1)*1002=-1002

A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )

A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 ) 

A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )

A = ( - 4 ) . 502 + 4019

A = - 2008 + 4019

A = 2011.

CHÚC LÀM BÀI VUI VẺ

Bài 1:

a)12,5 x (-5/7) + 1,5 x (-5/7)

=-5/7*(12,5+1,5)

=-5/7*14

=-10

b)(-1/4) x (6|2/11) + 3|9/11 x (-1/4)

=-1/4*(68/11+42/11)

=-1/4*10

=-5/2

c tương tự

d)\(\frac{9^8\cdot4^3}{27^4\cdot6^5}=\frac{\left(3^2\right)^8\cdot\left(2^2\right)^3}{\left(3^3\right)^4\cdot\left(2\cdot3\right)^5}=\frac{3^{16}\cdot2^6}{3^{12}\cdot2^5\cdot3^5}=\frac{3^{16}\cdot2^5\cdot2}{3^{16}\cdot3^1\cdot2^5}=\frac{2}{3}\)

Bài 2:

a)Ta có:

2⁸⁰⁰=(2⁸)¹⁰⁰=256¹⁰⁰

8²⁰⁰=(8²)¹⁰⁰=64¹⁰⁰

Vì 256¹⁰⁰>64¹⁰⁰ =>2⁸⁰⁰>8²⁰⁰

b)Ta có:

12⁴⁵=(12³)¹⁵=1728¹⁵

Vì 625¹⁵<1728¹⁵ =>625¹⁵<12⁴⁵

a) -5/7x(12,5+1,5)=-10

b) -1/4x (6I2/11+3I9/11) = -1/4x 10=-5/2

c) (-7/5 + -3/8 + -3/5+5/8): 2009/2010 = -7/4:2009/2010=-1005/574

d)3^16x 4^3/3^12x 6^5=3^4x4^3x6^5=......

bài 2)

2^800=2^4x200= 15^200> 8^200

=> 2^800>8^200

B) quên cách làm

dell   

bai o ban noi the co the ban la c ho day